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5 + 1 6 + 2 7 + 3 8 + 4 9 + 5 = ? \large \sqrt{5+1\sqrt{6+2\sqrt{ 7+3\sqrt{8+4\sqrt{ 9+5\sqrt{\dots}}}}}} = \ ?


This problem is a part of set nested radicals .

3 4 2 5 1

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1 solution

Ankith A Das
Mar 13, 2015

5 + 1 6 + 2 7 + 3 8 + 4 9 + 5 U s i n g R a m a n u j a n s n e s t e d r a d i c a l f o r m u l a x + n + a = a x + ( a + n ) 2 + x a ( x + n ) + ( a + n ) 2 + ( x + n ) . . . W e c a n s e e t h a t x = 1 t h e n x + n = 2 1 + n = 2 ; n = 1 S u b s t i t u t i n g x a n d n i n a x + ( a + n ) 2 a ( 1 ) + ( a + 1 ) 2 = 5 a = 1 x + n + a = 1 + 1 + 1 = 3 \sqrt { 5+1\sqrt { 6+2\sqrt { 7+3\sqrt { 8+4\sqrt { 9+5\sqrt { \dots } } } } } } \\ Using\quad Ramanujan's\quad nested\quad radical\quad formula\quad \\ x+n+a =\sqrt { ax+{ \left( a+n \right) }^{ 2 }+x\sqrt { a\left( x+n \right) +{ \left( a+n \right) }^{ 2 }+\left( x+n \right) \sqrt { ... } } } \\ We\quad can\quad see\quad that\quad x=1\\ then\quad x+n=2\\ 1+n=2\quad ;\quad n=1\\ Substituting\quad x\quad and\quad n\quad in\quad ax+{ \left( a+n \right) }^{ 2 }\\ a\left( 1 \right) +{ \left( a+1 \right) }^{ 2 }=\quad 5\\ a=1\\ x+n+a=1+1+1=3

We can do this without referring to Ramanujan's formula: Doing some numerical experimentation, we suspect that the answer is 3. We can prove this result by repeatedly using the formula x = x= ( x + 2 ) + ( x 2 ) ( x + 1 ) \sqrt{(x+2)+(x-2)(x+1)} , always applying it to the last number appearing in an expression: 3 = 5 + 1 4 3=\sqrt{5+1*4} = 5 + 1 6 + 2 5 \sqrt{5+1\sqrt{6+2*5}} = 5 + 1 6 + 2 7 + 3 6 \sqrt{5+1\sqrt{6+2\sqrt{7+3*6}}} =...

Otto Bretscher - 6 years, 2 months ago

brilliant!.. the level of this problem should be more than this..

Bhupendra Jangir - 6 years, 3 months ago

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true that.

Ankith A Das - 6 years, 3 months ago

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It is very easy to approximate the answer as 3, which is why the problem rating is so low.

Daniel Liu - 6 years, 3 months ago

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@Daniel Liu Can you please tell how you approximated the answer.

Ankith A Das - 6 years, 3 months ago

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@Ankith A Das Let's say that 6 + 2 . . . 9 6+2\sqrt{...}\approx 9 , so 5 + 6 + 2 . . . 8 5+\sqrt{6+2\sqrt{...}}\approx 8 so 5 + 6 + 2 . . . 3 \sqrt{5+\sqrt{6+2\sqrt{...}}}\approx 3

Daniel Liu - 6 years, 3 months ago

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